Question

Yzp^2-q=0 solve by charpit method

Answer 1

Given: $yzp^{2}-q=0$

To find:

• solution by Charpit's method

Solution:

Here the given equation is

$\quad F(x,y,z,p,q)=yzp^{2}-q=0$ .....(1)

∴ Charpit's auxiliary equations are

$\dfrac{dp}{\frac{\partial F}{\partial x}+p\frac{\partial F}{\partial z}}=\dfrac{dq}{\frac{\partial F}{\partial y}+q\frac{\partial F}{\partial z}}=\dfrac{dz}{-p\frac{\partial F}{\partial p}-q\frac{\partial F}{\partial q}}\\=\dfrac{dx}{-\frac{\partial F}{\partial p}}=\dfrac{dy}{-\frac{\partial F}{\partial q}}$

$\Rightarrow \dfrac{dp}{0+p.yq^{2}}=\dfrac{dq}{zp^{2}+q.yp^{2}}\\=\dfrac{dz}{-p.2yzp-q.(-1)}=\dfrac{dx}{-2yzp}=\dfrac{dy}{-1}$

$\Rightarrow \dfrac{dp}{yp^{3}}=\dfrac{dq}{zp^{2}+yp^{2}q}=\dfrac{dz}{-2yzp^{2}+q}\\=\dfrac{dx}{-2yzp}=\dfrac{dy}{1}$

Taking the first and the last fractions, we get

$\quad \dfrac{dp}{yp^{3}}=\dfrac{dy}{1}$

$\Rightarrow \dfrac{dp}{p^{3}}=ydy$

Integrating we get

$\quad \int \dfrac{dp}{p^{3}}=\int ydy$

$\Rightarrow -\dfrac{1}{2p^{2}}=\dfrac{y^{2}}{2}-\dfrac{c^{2}}{2}$

where $c$ is constant of integration

$\Rightarrow \dfrac{1}{p^{2}}=c^{2}-y^{2}$

$\Rightarrow p^{2}=\dfrac{1}{c^{2}-y^{2}}$

$\Rightarrow \boxed{p=\dfrac{1}{\sqrt{c^{2}-y^{2}}}}$

Then $\boxed{q=\dfrac{yz}{c^{2}-y^{2}}}$ by (1)

Substituting the values of $p$ and $q$ in

$\quad dz=pdx+qdy$, we get

$\quad dz=\dfrac{1}{\sqrt{c^{2}-y^{2}}}+\dfrac{yz}{\sqrt{c^{2}-y^{2}}}dy$

$\Rightarrow dz=\dfrac{1}{\sqrt{c^{2}-y^{2}}}[dx+\dfrac{yzdy}{\sqrt{c^{2}-y^{2}}}$

$\Rightarrow \sqrt{c^{2}-y^{2}}dz=dx+\dfrac{yzdy}{\sqrt{c^{2}-y^{2}}}$

$\Rightarrow \sqrt{c^{2}-y^{2}}dz-\dfrac{yzdy}{\sqrt{c^{2}-y^{2}}}=dx$ .....(2)

Let $\sqrt{c^{2}-y^{2}}z=t$ so that

$\quad \sqrt{c^{2}-y^{2}}dz-\dfrac{yzdy}{\sqrt{c^{2}-y^{2}}}=dt$

Continuing (2), we get

$\quad dt=dx$

$\Rightarrow t=x+k$ ( by integration )

where $k$ is constant of integration

$\Rightarrow \sqrt{a^{2}-y^{2}}z=x+k$

Squaring we get

$\quad (a^{2}-y^{2})z^{2}=(x+k)^{2}$

This is the required solution.

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